# Showing a matrix is diagonalizable and calculating powers

## Homework Statement

Consider the matrix A =
| 7 16 8|
|-1 0 -1|
|-2 -10 -3|

Show that A is diagonalizable. Find an invertible matrix P and diagonal matrix D and use the obtained result in order to calculate A^2 and A^3

## Homework Equations

Determinant equation I suppose.
D is the diagonal matrix, like identity matrix but constructed from the eigenvalues. The order of the eigenvalues must match the columns in P precisely.

## The Attempt at a Solution

Alright well I've done a huge chunk of this problem already. I found the eigenvalues to be 2, 3, and -1. Then I find my eigenvectors, and I come up with 2 per eigenvalue.

For λ = 2 I get [-2 1 0] and [-1 0 1]
For λ = 3 I get [-3 1 0] and [-1 0 1]
For λ = 1 I get [1 1 0] and [0 1 1]

So now I have to construct P from these. This is where I'm confused. Which eigenvectors from which eigenvalues do I use? I've tried several combinations to make AP = PD and I just can't do it. Everything comes out wrong. I suppose there's just something I'm not understanding. Please help.

Mark44
Mentor

## Homework Statement

Consider the matrix A =
| 7 16 8|
|-1 0 -1|
|-2 -10 -3|

Show that A is diagonalizable. Find an invertible matrix P and diagonal matrix D and use the obtained result in order to calculate A^2 and A^3

## Homework Equations

Determinant equation I suppose.
D is the diagonal matrix, like identity matrix but constructed from the eigenvalues. The order of the eigenvalues must match the columns in P precisely.

## The Attempt at a Solution

Alright well I've done a huge chunk of this problem already. I found the eigenvalues to be 2, 3, and -1. Then I find my eigenvectors, and I come up with 2 per eigenvalue.

For λ = 2 I get [-2 1 0] and [-1 0 1]
For λ = 3 I get [-3 1 0] and [-1 0 1]
For λ = 1 I get [1 1 0] and [0 1 1]

So now I have to construct P from these. This is where I'm confused. Which eigenvectors from which eigenvalues do I use? I've tried several combinations to make AP = PD and I just can't do it. Everything comes out wrong. I suppose there's just something I'm not understanding. Please help.
Since you have a 3 x 3 matrix and found three eigenvalues, you should not have two eigenvectors per eigenvalue, so you have definitely done something wrong.

Once you have that squared away, it's a good idea to check your eigenvalues and eigenvectors. If λ is an eigenvalue, and x is an eigenvector associated with that eigenvalue, it should be true that Ax = λx. If not, you did something wrong.

If you end up with three eigenvectors, form a matrix P with those eigenvectors as columns. It doesn't matter what order you put them in, but the order will determine which eigenvalue goes where in your diagonal matrix D.

HallsofIvy
Homework Helper

## Homework Statement

Consider the matrix A =
| 7 16 8|
|-1 0 -1|
|-2 -10 -3|

Show that A is diagonalizable. Find an invertible matrix P and diagonal matrix D and use the obtained result in order to calculate A^2 and A^3

## Homework Equations

Determinant equation I suppose.
D is the diagonal matrix, like identity matrix but constructed from the eigenvalues. The order of the eigenvalues must match the columns in P precisely.

## The Attempt at a Solution

Alright well I've done a huge chunk of this problem already. I found the eigenvalues to be 2, 3, and -1. Then I find my eigenvectors, and I come up with 2 per eigenvalue.

For λ = 2 I get [-2 1 0] and [-1 0 1]
For λ = 3 I get [-3 1 0] and [-1 0 1]
For λ = 1 I get [1 1 0] and [0 1 1]

So now I have to construct P from these. This is where I'm confused. Which eigenvectors from which eigenvalues do I use? I've tried several combinations to make AP = PD and I just can't do it. Everything comes out wrong. I suppose there's just something I'm not understanding. Please help.
$$\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}-2 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}2 \\ 2 \\ -6\end{bmatrix}$$
so this is NOT an eigenvector.

$$\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix}= \begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix}$$
so this is and eigenvector but with eigenvalue -1, not 2.

$$\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}-3 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}-5 \\ 3 \\ -4\end{bmatrix}$$
so this is not an eigenvector.

[-1 0 1] is the same vector as in the second case. It is an eigenvector with eigenvalue -1, not 3.

$$\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}23 \\ -1 \\ -12\end{bmatrix}$$
so this is not an eigenvector.

$$\begin{bmatrix} 7 & 16 & 8 \\ -1 & 0 & -1 \\-2 & -10 & -3\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix}= \begin{bmatrix}24 \\ -1 \\ -13\end{bmatrix}$$
so this is not an eigenvector.

Last edited by a moderator:

I did the work again and found my mistake. The eigenvectors are

[0 -1 2], [2 -1 1], [-1 0 1]